Question

At a pressure of $760torr$ and temperature of $273.15K$ the indicated volume of which system is not consistent with the observation
A) $14g\left( {{N}_{2}} \right)+16g({{O}_{2}}):volume=22.4L$
B) $4g\left( He \right)+44g(C{{O}_{2}}):volume=44.8L$
C) $7g\left( {{N}_{2}} \right)+36g({{O}_{3}}):volume=22.4L$
D) $17g\left( N{{H}_{3}} \right)+36.5g(HCl):volume=44.8L$

Answer 1

We know that in the given question the volume of the subsequent system has been observed. We have to calculate the following equations and observe the correct one. And by knowing a gas molar volume at a particular temperature and a particular pressure can give the calculation of the volume engaged by any number of moles of that corresponding gas.

Complete answer:
We have $Pressure=1atm$ and $temperature=73.15K$. We can calculate the mole of compounds by adding them as $1mol$ of nitrogen $\left( {{N}_{2}} \right)$ has a mass of $14g$ (the mass on the periodic table). And $1mol$ of oxygen $({{O}_{2}})$ has a mass of $16g$ from the mass on the periodic table. Now, for calculation of volume, we know
$1$ mole of any gaseous mixture $=22.4L$ So, $n$ mole of any gaseous mixture $=n\times 22.4L$

So, the Volume of mixture $=n\times 22.4$ and which can be given as:
A) $14g\left( {{N}_{2}} \right)+16g~\left( {{O}_{2}} \right)=\dfrac{14}{28}+\dfrac{16}{32}=\dfrac{1}{2}+\dfrac{1}{2}\text{ }=1mole$
So, the Volume $=22.4L$ ………………since, Volume of mixture is given by $n\times 22.4$

B) $~4g\left( He \right)+44g~\left( C{{O}_{2}} \right)=\dfrac{4}{4}+\dfrac{44}{44}=1+1=2mole$
So, the Volume $=2\times 22.4=44.8L$ ………………since, Volume of mixture is given by $n\times 22.4$

C) $~7g\left( {{N}_{2}} \right)+36g\left( {{O}_{3}} \right)=\dfrac{7}{28}+\dfrac{36}{48}=\dfrac{1}{4}+\dfrac{3\text{ }}{4}=1mole$
So, the Volume $=22.4L$ ………………since, Volume of mixture is given by $n\times 22.4$

D) $17g\left( N{{H}_{3}} \right)+36.5g\left( HCl \right)=\dfrac{17}{17}+\dfrac{36.5\text{ }}{36.5}=2mole$

So, the Volume of mixture $=2\times 22.4=44.8L$ ………………since, Volume of mixture is given by $n\times 22.4$
Since, here reaction is $N{{H}_{3}}{{}_{\left( g \right)}}+HC{{l}_{\left( g \right)}}\to N{{H}_{4}}C{{l}_{\left( s \right)}}$ So, the Volume of solid is more than the mixture. Therefore, Volume of mixture is $<<44.8L$

Therefore correct answer is option D i.e. at pressure of $760torr$ along with temperature of $273.15K$ indicated volume of which system is not consistent with observation is $17g\left( N{{H}_{3}} \right)+36.5g(HCl):volume=44.8L$.

Hence the correct answer is option ‘D’.

Note: Note that the mole can be defined as the amount of substance. It can be expressed as grams, molecules, particles, liters, atoms. The molar volume of a gas states the volume occupied by $1$ mole of that particular gas under convinced temperature and pressure conditions.